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Burka [1]
2 years ago
13

A homogenous mixture that is mainly composed of water is referred to as a(n) __________ solution.

Chemistry
1 answer:
rosijanka [135]2 years ago
7 0

Answer:

solvent solutions

Explanation:

.............

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Which of the five type of equilibrium problems best applies to this question: Consider the following reaction at equilibrium. Wh
andreev551 [17]

Answer:

d

Explanation:

3 0
3 years ago
Passing an electric current through a sample of water (H2O) can cause the water to decompose into hydrogen gas (H2) and oxygen g
Andreas93 [3]
2H2O --> 2H2 + O2
The mole H2O:mole O2 ratio is 2:1
Now determine how many moles of O2 are in 50g: 50g × 1mol/32g = 1.56 moles O2
Since 1 mole of O2 was produced for every 2 moles of H2O, we need 2×O2moles = H2O moles
2×1.56 = 3.13 moles H2O
Finally, convert moles to grams for H2O:
3.13moles × 18g/mol = 56.28 g H2O
D) 56.28
7 0
3 years ago
In the absence of sodium methoxide, the same alkyl bromide gives a different product. Draw an arrowpushing mechanism to account
hoa [83]

Answer:

See explanation below

Explanation:

The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.

Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.

For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)

For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.

3 0
3 years ago
2C 2 H 6 +7O 2 ***>4CO 2 +6H 2 O if 7.0 g of C 2 H 6 react with 18 g of O 2 , how many grams of water will be produced
Alex787 [66]

Answer:

grams H₂O produced = 8.7 grams

Explanation:

Given 2C₂H₆(g) + 7O₂(g) => 4CO₂(g) + 6H₂O(l)

               7g           18g                             ?g

Plan => Convert gms to moles => determine Limiting reactant => solve for moles water => convert moles water to grams water

Moles Reactants

moles C₂H₆ = 7g/30g/mol = 0.233mol

moles O₂ = 18g/32g/mol = 0.563mol

Limiting Reactant => (Test for Limiting Reactant)  Divide mole value by respective coefficient of balanced equation; the smaller number is the limiting reactant.

moles C₂H₆/2 = 0.233/2 = 0.12

moles O₂/7 = 0.08

<u><em>Limiting Reactant is O₂</em></u>

Moles and Grams of H₂O:

Use Limiting Reactant moles (not division value) to calculate moles of H₂O.

moles H₂O = 6/7(moles O₂) = 6/7(0.562) moles H₂O = 0.482 mole H₂O yield

grams H₂O = (0.482mol)(18g·mol⁻¹) = 8.7 grams H₂O

3 0
3 years ago
What are some examples of gases at room temperature
Lady_Fox [76]

Answer:

Nitrogen, Hydrogen, Oxygen, Chlorine, and Fluorine are all gases at room temperature.

Explanation:

6 0
3 years ago
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